Express $\int^d_c \frac{\mathrm{d}u}{\sqrt{b^2\sinh^2 u -a}}du$ in terms of elliptic integral.

Question

I want to express $$\int^d_c \frac{\mathrm{d}u}{\sqrt{b^2\sinh^2 u -a}}du$$ (where $a,b,c,d$ are real constants) in terms of elliptic functions. I have tried to make the naive change of variable $b^2\sinh^2 u -a =1-k^2\sin^2\theta$ (where $k$ is to be determined in terms of $a,b,c,d$ so the integral simplifies), but its Jacobian factor seemingly introduces more complication that simplification. Does anybody have any suggestion on how to proceed?

Answer 1

Consider the indefinite integral in general:

$$I=\int \frac{1}{\sqrt{b^2\sinh^2 u -a}} du$$

Let $s^2=\frac{a}{b^2}, t=\sinh u, du=\frac{1}{\sqrt{1+t^2}}dt$

$$I=\frac{1}b\int \frac{1}{\sqrt{t^2-s^2}}\frac{1}{\sqrt{1+t^2}}dt$$

Let $t=\cot\theta$

$$I=-\frac{1}b\int \frac{1}{\sqrt{\cos^2\theta-s^2\sin^2\theta}}d\theta$$

Let $k^2=1+s^2$ and use: $\cos^2\theta=1-\sin^2\theta$. We have

$$I=-\frac{1}b\int \frac{1}{\sqrt{1-k^2\sin^2\theta}}d\theta$$

So you get the elliptic integral of the first kind.